Find the surface area of the surface z=cosh(sqrt(x^2+y^2)) above the region in the xy plane given in polar coordinates:
r is between 0 and theta
theta is between 2 and 4

Ok. I used the formula:
Surface area equals the square root of the partial derivative of x squared plus the partial derivative of y squared plus 1. After doing it out, the (x^2+y^2) turned into 1, leaving me with the square root of 1 + (sinhsqrt(x^2+y^2))^2.
I took the integral of this from 0 to theta by switching the (x^2+y^2) into just r (polar coordinates). I got that this integral equaled cosh(r)tanh(r) and in turn cosh(theta)tanh(theta). Taking this integral from 2 to 4, I got cosh(4)-cosh(2). I thought I did everything correct, but this is not the answer. What did I do wrong?

Thanks for any help. I hope you can understand my thought process.

Aug 2nd 2006, 10:16 AM

ThePerfectHacker

Quote:

Originally Posted by MikeWakefield32

r is between 0 and theta

You mean,? :confused:

That is not possible, cuz the region is unbounded!
(The curve produces an Archimedean Spiral).

Aug 2nd 2006, 10:18 AM

MikeWakefield32

1theta it says. I didn't think that mattered.

Aug 2nd 2006, 10:46 AM

ThePerfectHacker

My fault the region is bounded. Because you gave theta a limitation. Sorry.
---
Then the integration is,
Where, is the region you described. Since is simpler expressed in polar coordinates we will use polar integration.
---
The surface is,
Then,
Then,
The next part is two square each one,
Add them,
Finally add 1,
But,
Therefore,
Thus, the integration is,
This simplifies as,
But, thus,
Now, finally you can use polar substituion,
Can you solve it from here?

Aug 2nd 2006, 11:14 AM

ThePerfectHacker

Quote:

Originally Posted by ThePerfectHacker

Can you solve it from here?

Let me continue.

To solve this integral the problem is to find,
You gonna use integration by parts.
Let, and
Then, and
Using parts,
Thus,
Thus, is,
Thus,
The problem now is to find,
Let, and
Then, and
Thus, by parts,
Thus,
Thus, you have, (entire integral)